What is the integral from 0 to 10 of (6e^{0.2x}+2) ?

The integral from 0 to 10 of (6e^{0.2x}+2) is 30e^2-10

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integral from 0 to 10 of (6e^{0.2x}+2)

\int_{0}^{10} (6 e^{0.2 x} + 2) d x

30 e^{2} - 10

Decimal Notation

211.67168 \dots

Solution steps

\int_{0}^{10} 6 e^{0.2 x} + 2 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{10} 6 e^{0.2 x} d x + \int_{0}^{10} 2 d x

\int_{0}^{10} 6 e^{0.2 x} d x = 30 (e^{2} - 1)

\int_{0}^{10} 2 d x = 20

= 30 (e^{2} - 1) + 20

Simplify

30 (e^{2} - 1) + 20 : 30 e^{2} - 10

= 30 e^{2} - 10

\int \frac{3 x}{2 x} d x

x \to 4 + lim (\frac{6 x}{x - 4})

f (x) = ln (x^{8})

\frac{d}{d y} (arctan (x y))

What is the integral from 0 to 10 of (6e^{0.2x}+2) ?
The integral from 0 to 10 of (6e^{0.2x}+2) is 30e^2-10